1. Optimization Mechanics of the Simplex Method

2. Simplex Method Find initial corner point (basic) feasible solution.
Usually, the origin is a basic feasible solution.
Iterate until the stopping conditions are met.
Are we optimal yet? Look at the current version of the objective function to see if an entering basic variable is available. If not, then the current basic feasible solution is the optimum solution.
Select the entering variable: choose the non-basic variable that gives the fastest rate of increase in the objective function value.
Select the leaving variable by applying the Minimum Ratio Test.
Update the equations to reflect new basic feasible solution.
Go to step (a).

3. Tableau
Basic Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
-175
never 7 11 77 10 8 80 9 6
First, we create a "tableau" out of the augmented system of equations and pick the origin as the starting point.

4. Tableau (Proper Form) Characteristics of a tableau in proper form:
Basic Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
-175
never 7 11 77 10 8 80 9 6
Characteristics of a tableau in proper form:
Exactly 1 basic variable per equation.
The coefficient of the basic variable is always exactly +1, and the coefficients above and below the basic variable in the same column are all 0.
Z is treated as the basic variable for the objective function row.

5. Tableau (Proper Form) When a tableau is in the proper form
Basic Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
-175
never 7 11 77 10 8 80 9 6
When a tableau is in the proper form
The RHS values are the value of the basic variables.
What is the value of x1?
x1 is a non-basic variable, so its value is 0.
What is the value of Z?
0. (Z is a basic variable, so look at the RHS column)

6. Are we optimal yet?
Basic Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
-175
never 7 11 77 10 8 80 9 6
We are optimal if no entering basic variable is available (i.e., no basic variable, if introduced, can increase the value of Z).
If there is any negative coefficient in the objective function row, then the current basic solution is not optimal.

7. Select the entering basic variableZ x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
-175
never 7 11 77 10 8 80 9 6
Select the variable with the most negative coefficient in the objective function row as the entering variable.
We won't accidentally select a basic variable. Why?
The column corresponds to the entering basic variable becomes the pivot column.

8. Select the leaving basic variableZ x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
-175
never 7 11 77
77/11=7 10 8 80
80/8=10 9
no limit 6
6/1=6
Perform Minimum Ratio Test (MRT)
Special cases: If the coefficient of the entering basic variable is not a positive number, enter "no limit".
Select the variable corresponding to the smallest MRT value as the leaving basic variable. Call the corresponding row the pivot row.

9. Updating the tableau
Basic Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
-175
never 7 11 77 10 8 80 9 6
Step 1: Replace the leaving basic variable in the "Basic" column by the entering basic variable.
Step 2: Turn the tableau into a proper form
How?

10. Updating the tableau
Basic Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
-175
never 7 11 77 10 8 80 9 6
The coefficient of all the basic variables must be +1.
Divide the row corresponds to the entering basic variable by its coefficient.
The coefficients above and below the basic variable in the same column must be turned into 0.
Apply (Gaussian) elimination steps

11. End of iteration #1 When x1 = 0, x2 = 6, Z = 1050.
Basic Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
175
1050
never 7
-11 11 10 -8 32 9 6
When x1 = 0, x2 = 6, Z = 1050.
Note: The columns correspond to the "old" basic variables do not change. (Why?)
Are we optimal yet?

12. Iteration #2 Select the entering basic variable: x1Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
175
1050
never 7
-11 11
11/7=1.571 10 -8 32
32/10=3.2 9 6
no limit
Select the entering basic variable: x1
Perform MRT test and select the leaving basic variable: S1

13. Iteration #2 Updating tableau Replace S1 by x1 in the "Basic" columnZ x1 x2 S1 S2 S3 S4
RHS
MRT 1
-150
175
1050
never
1/7
-11/7
11/7 10 -8 32 9 6
Updating tableau
Replace S1 by x1 in the "Basic" column
Divide pivot row by coefficient of x1 (=7)

14. Iteration #2 Updating tableau (continue) (End of iteration #2)
Basic Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
150/7
-425/7
9000/7
never
1/7
-11/7
11/7
-10/7
54/7
114/7
-1/7
52/7 6
Updating tableau (continue)
Apply elimination steps.
(End of iteration #2)
Are we optimal yet?

15. Iteration #3 Select the entering basic variable: S4Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
150/7
-425/7
9000/7
never
1/7
-11/7
11/7
no limit
-10/7
54/7
114/7
114/54
-1/7
52/7
52/11 6
Select the entering basic variable: S4
Perform MRT test and select the leaving basic variable: S2

16. Iteration #3 Updating tableau Replace S2 by S4 in the "Basic" columnZ x1 x2 S1 S2 S3 S4
RHS
MRT 1
150/7
-425/7
9000/7
never
1/7
-11/7
11/7
-10/54
7/54
114/54
-1/7
52/7 6
Updating tableau
Replace S2 by S4 in the "Basic" column
Divide pivot row by coefficient of S4 (=54/7)

17. Iteration #3 Updating tableau (continue) (End of iteration #3)
Basic Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
10.185
7.870
never
-0.148
0.204
4.889
-0.185
0.130
2.111
0.148
-0.204
4.111
0.185
-0.130
3.889
Updating tableau (continue)
Apply elimination steps.
(End of iteration #3)
Are we optimal yet?

18. Iteration #3
Basic Z x1 x2 S1 S2 S3 S4
RHS
MRT 1
10.185
7.870
never
-0.148
0.204
4.889
-0.185
0.130
2.111
0.148
-0.204
4.111
0.185
-0.130
3.889
Because there is no more negative coefficient in the objective function row, we cannot further improve the value of the objective function.
The optimal value of the objective function is when x1 = and x2 =

19. Special Cases in Tableau Manipulation
In case of tie for the entering basic variable, select any of the tying variables.
When will a tie occur for the entering variable?
In case of tie for the leaving basic variable, select any of the tying variables.
When will a tie occur for the leaving variable?
What does it mean when all MRT yield "No limit"?
Increasing the value of the entering variable does not affect any variable. i.e., the problem is unbounded.

20. Special Cases in Tableau Manipulation
Note: Here the coefficient is referring to the coefficient of a variable in the objective function row in the tableau.
At the optimum, what's the coefficient of a non-basic variable?
At the optimum, what's the coefficient of a basic variable?
What does it mean when the coefficient of a non-basic variable is zero at the optimum?
We have alternate optimal solutions. i.e., increasing the value of that non-basic variable does not increase or decrease the value of Z.

21. Directions & References
Solving non-standard linear programming.
Equality and greater-than-or-equal-to (≥) constraints
Variables that can be negative or unrestricted variables
Sensitivity Analysis
LP in practice: Revised Simplex method, interior point methods.
Ref:

22. Summary Simplex Method for solving "Standard Form" LP problems
Algorithm
Understand the characteristics of the slack variables
Understand how to select entering/leaving variables
Performing simplex method in tableau form
Understand the characteristics of the tableau (which can give you additional info of the problem).

23. Summary – Optimization
One dimensional unconstrained optimization – function with one variable
Newton's Method for finding x s.t. f'(x) = 0.
Bracketing methods (Iterative)
Golden Section Search
Quadratic Interpolation
Hybrid methods

24. Summary – Optimization
Multi-dimensional unconstrained optimization – function with 2 or more variables
Random Search
Its advantages and disadvantages
The "General Iterative Algorithm" for finding optimal point using either non-gradient and gradient methods.
Non-gradient Methods
Univariate Search and Powell's Method
Gradient Methods
Evaluate partial derivative
Derive gradients and Hessian matrix
Determine if a point is a maximum, minimum, or saddle point using Hessian matrix
Finding optimal point in a given direction
Steepest Ascent and Newton's Method